How do I use bash to compare 1000 files in sequential order - 1 to 2, 2 to 3, and so on-CodePudding

I have about 1000 html files in a directory that are different versions of a webpage. Each is named by the date the page was modified - e.g. 20190104, 20190105 - but there are occasional gaps, such as between 20190105 and 20190107.

I know how to use diff to compare one file to the immediately subsequent version and save the differences to a third file, but I want a loop that will that for all 1000. The standard "for i in ./*" doesn't work because I need to reference multiple files inside the loop.

I suspect I need to turn the list of files into an array, and then have a for loop that references array elements n and n 1, but I'm having trouble with the syntax.

CodePudding user response：

elements n and n 1

No, don't think in the future - you don't know it. You know the past - iterate over n-1 and n.

prev=  # empty previous filename
for cur in ./*; do
   if [[ -n "$prev" ]]; then    # if prev is Nonempty
        do_stuff "$prev" "$cur"
   fi
   prev=$cur
done

CodePudding user response：

No need for an array

last=BEGIN
for f in $(echo * | sort); do 
    echo diff "$last" "$f" 
    diff "$last" "$f" 
    last="$f"
done

CodePudding user response：

Comparing by pairs:

#!/usr/bin/env bash

comparePairs() {
  while [ "$2" ]; do
    printf 'Compare %s with %s.\n' "$1" "$2"
    shift
  done
}

comparePairs foo bar baz cuux

output:

Compare foo with bar.
Compare bar with baz.
Compare baz with cuux.

Now your diff pairs:

#!/usr/bin/env bash

diffPairs() {
  while [ "$2" ]; do
    diff "$1" "$2" > "$1-$2.diff"
    shift
  done
}

diffPairs "$@"